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Question Number 160597 by cortano last updated on 03/Dec/21

$${\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dx}}{2-\cos \mathrm{x}}=?$$

Answered by Mathspace last updated on 03/Dec/21

$$I={\int }_0^{\frac{\pi }{2}}\frac{dx}{2-cosx}wedothechangement$$$$tan\left(\frac{x}{2}\right)=t\Rightarrow I={\int }_0^1\frac{2dt}{(1+t^2)(2-\frac{1-t^2}{1+t^2})}$$$$=2{\int }_0^1\frac{dt}{2+2t^2-1+t^2}=2{\int }_0^1\frac{dt}{3t^2+1}$$$${=}_{\sqrt{3}t=u}2{\int }_0^{\sqrt{3}}\frac{du}{\sqrt{3}(1+u^2)}$$$$=\frac{2}{\sqrt{3}}{\left[arctanu\right]}_0^{\sqrt{3}}=\frac{2}{\sqrt{3}}arctan\left(\sqrt{3}\right)$$

Commented by cortano last updated on 03/Dec/21

$$\mathrm{yes}....$$

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